The centre of the mass of \(3\) particles, \(10\) kg, \(20\) kg, and \(30\) kg, is at \((0,0,0)\). Where should a particle with a mass of \(40\) kg be placed so that its combined centre of mass is \((3,3,3)\)?
1. \((0,0,0)\)
2. \((7.5, 7.5, 7.5)\)
3. \((1,2,3)\)
4. \((4,4,4)\)

The position of a particle is given by \(\vec r = \hat i+2\hat j-\hat k\) and momentum \(\vec P = (3 \hat i + 4\hat j - 2\hat k)\). The angular momentum is perpendicular to:

A wheel with a radius of \(20\) cm has forces applied to it as shown in the figure. The torque produced by the forces of \(4\) N at \(A\), \(8~\)N at \(B\), \(6\) N at \(C\), and \(9~\)N at \(D\), at the angles indicated, is:

1. \(5.4\) N-m anticlockwise

2. \(1.80\) N-m clockwise

3. \(2.0\) N-m clockwise

4. \(3.6\) N-m clockwise

A particle of mass \(m\) moves in the\(XY\) plane with a velocity of \(v\) along the straight line \(AB.\) If the angular momentum of the particle about the origin \(O\) is \(L_A\) when it is at \(A\) and \(L_B\) when it is at \(B,\) then:

A wheel is rotating about an axis through its centre at \(720\) r.p.m. It is acted upon by a constant torque opposing its motion for \(8\) seconds to bring it to rest finally. The value of torque in N-m is: (given \(I\) = $\frac{24}{\mathrm{\pi }}$ kg ${\mathrm{m}}^2$)
1. \(48\)
2. \(72\)
3. \(96\)
4. \(120\)

For L = 3.0 m, the total torque about pivot A provided by the forces as shown in the figure is:

Two rotating bodies \(A\) and \(B\) of masses \(m\) and \(2m\) with moments of inertia $I_A$ and $I_B (I_B>I_A)$ have equal kinetic energy of rotation. If $L_A$ $$and $L_B$ be their angular momenta respectively, then:
1. $L_A=\frac{L_B}{2}$
2. $L_A=2L_B$
3. $L_B>L_A$
4. $L_A>L_B$

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through:
 
1. B                   
2. C
3. D                   
4. A